poj 2096 Collecting Bugs （概率与期望DP）

Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 5242   Accepted: 2598 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. Each day he discovers exactly one bug in the program and adds information about it and its category into a spreadsheet. When he finds bugs in all bug categories, he calls the program disgusting, publishes this spreadsheet on his home page, and forgets completely about the program.  Two companies, Macrosoft and Microhard are in tight competition. Microhard wants to decrease sales of one Macrosoft program. They hire Ivan to prove that the program in question is disgusting. However, Ivan has a complicated problem. This new program has s subcomponents, and finding bugs of all types in each subcomponent would take too long before the target could be reached. So Ivan and Microhard agreed to use a simpler criteria --- Ivan should find at least one bug in each subsystem and at least one bug of each category.  Macrosoft knows about these plans and it wants to estimate the time that is required for Ivan to call its program disgusting. It's important because the company releases a new version soon, so it can correct its plans and release it quicker. Nobody would be interested in Ivan's opinion about the reliability of the obsolete version.  A bug found in the program can be of any category with equal probability. Similarly, the bug can be found in any given subsystem with equal probability. Any particular bug cannot belong to two different categories or happen simultaneously in two different subsystems. The number of bugs in the program is almost infinite, so the probability of finding a new bug of some category in some subsystem does not reduce after finding any number of bugs of that category in that subsystem.  Find an average time (in days of Ivan's work) required to name the program disgusting. Input Input file contains two integer numbers, n and s (0 < n, s <= 1 000). Output Output the expectation of the Ivan's working days needed to call the program disgusting, accurate to 4 digits after the decimal point. Sample Input 1 2 Sample Output 3.0000 Source Northeastern Europe 2004, Northern Subregion

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题目大意：一个软件有s个子系统，会产生n种bug某人一天发现一个bug,这个bug属于一个子系统，属于一个分类每个bug属于某个子系统的概率是1/s,属于某种分类的概率是1/n
问发现n种bug,每个子系统都发现bug的天数的期望。
题解：概率与期望DP

dp[i][j]表示出现i种bug,j个系统发现bug的期望，有四种转移可以转移到dp[i][j]

dp[n][s]=0,我们要求的就是dp[0][0]

dp[i][j]->dp[i][j] 概率是i/n*j/s

dp[i][j+1]->dp[i][j]  i/n*(1-j/s)

dp[i+1][j]-> dp[i][j] (1-i/n)*j/s

dp[i+1][j+1]-> dp[i][j] (1-i/n)*(1-j/s)

将上面的式子进行整理，把dp[i][j]当成未知项，移项，解出来就可以了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define N 1003
using namespace std;
int n,s;
double dp[N][N];
int main()
{
	scanf("%d%d",&n,&s);
	dp[n][s]=0;
	for (int i=n;i>=0;i--)
	 for (int j=s;j>=0;j--) {
	 	if (i==n&&j==s) continue;
	 	double p=1.0-i*1.0/n;
	 	double q=1.0-j*1.0/s;
	 	double p1=i*1.0/n;
	 	double q1=j*1.0/s;
	 	dp[i][j]=((dp[i+1][j+1]+1)*p*q+(dp[i+1][j]+1)*p*q1+(dp[i][j+1]+1)*p1*q+p1*q1)/(1-p1*q1);
	 }
	printf("%.4lf\n",dp[0][0]);
}